I am trying to show the system $$x'=y-\frac{xf(r)}{r}, \ y'=-x-\frac{yf(r)}{r}$$ where $r^2=x^2+y^2$, has isolated periodic solutions (i.e. limit cycles) corresponding to the real zeros of $f(r)$. Determine all the limit cycles for the case $$f(r)=r(r-2)^2(r^2-4r+3).$$
Using polar coordinates ($x=r\cos(\theta), \ y=r\sin(\theta)$, it is easy to show that $$r'=-f(r)=-r(r-2)^2(r^2-4r+3)=-r(r-2)^2(r-1)(r-3).$$ Hence $r'=0\implies r=0,1,2,3$. It can also be shown that $$\theta'=-1.$$
I can see that $r=1,2,3$ correspond to periodic solutions (as $r>0$ and $\theta'$ is fixed). However, how do we know that $r=0$ corresponds to a fixed point? Is this simply because \begin{align} r=0\implies x=y=0\implies x'&=y'=0? \end{align} An intuitive or even a geometric answer would be very appreciated.
Another motivating example (Perko Differential Equations and Dynamical Systems Third Edition, page 205)
$$x'=-y+x(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right),$$ $$x'=x+y(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right).$$
Using polar coordinates, it can be shown that $$r'=r^3\sin\left(\frac{1}{r}\right).$$ How do we know that $r=0$ corresponds to a fixed point at the origin (this is just explicitly stated)? To me this make counterintuitive sense, as I thought $r=0\implies x=y=0$, which clearly isn't allowed in this system.